2009 AIME II Problems/Problem 3

Solution

$[asy] pair A=(0,10), B=(0,0), C=(14,0), D=(14,10); draw (A--B--C--D--cycle); pair E=(7,10); draw (B--E); draw (A--C); pair F=(6.7,6.7); label("$E$",E,N); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$F$",F,W); [/asy]$

From the problem, $AB=100$ and triangle $AFB$ is a right triangle. As $ABCD$ is a rectangle, triangles $ADB$ , and $ABE$ are also right triangles. By $AA$ , $\triangle AFB \sim \triangle ADB$ , and by the altitude-on-hypotenuse theorems, $\triangle AFB \sim \triangle ABE$ , so $\triangle ABE \sim \triangle ADB$ . This gives $\frac {AE}{AB}= \frac {AB}{BC}$ . $AE=\frac{AD}{2}$ and $BD=AD$ , so $\frac {AD}{2AB}= \frac {AB}{AD}$ , or $(AD)^2=2(AB)^2$ , so $AD=AB \sqrt{2}$ , or $100 \sqrt{2}$ , so the answer is $\boxed{141}$ .

Page

Toolbox

Search

2009 AIME II Problems/Problem 3

Solution